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treat y as a function of x so the derivative with respect to x, of y^2 is 2y.dy/dx use the product rule to differentiate 4xy find the derivative term by term then make dy/dx the subject
does this help?
I get Y -4dy/dx for the -4xy. is this correct?
no its -(4x * dy/dx + 4y) = -4x dy/dx - 4y
so we have 2x - 4x.dy/dx - 4y + 2y.dy/dx = 0 now solve for dy/dx
I get Y Prime = (x+3y)/ -2x I know this is not the answer
I should be getting (2y-x)/(y-2x)
But I'm not.
because I know the answer, but that does me no good if I don't understand how to get the answer. The part that is tripping me up the most is that -4xy. we use y' instead of dy/dx in our class.
2x - 4x.dy/dx - 4y + 2y.dy/dx = 0 2y.dy/dx - 4x dydx = 4y - 2x dy/dx = (4y - 2x) / ( 2y - 4x) dividing top and bottm by 2 dy/dx = (2y - x)/(y - 2x)
can you explain how you get to your answer
derivative of -4xy = derivative of - (4xy) = - ( 4x.y' + y* 4) = -4xy' - 4y
Product Rule: f'(xy) = f(x)f'(y) + f(y)f'(x)
the derivative of y^2 is found using the chain rule
it's mainly the algebra that is messing me up . being able to get 2x - 4xy'-4y=2yy' into the form (2y-x)/ (y-2x)
I am having trouble getting the Y' by itself on one side.
2y.dy/dx - 4x dydx = 4y - 2x = 2y y' - 4x y' = 4y - 2x factor the left side (y' is common to both terms):- y' (2y - 4x) = 4y - 2x Now divide both sides by 2y-4x:- y' = (4y - 2x)/(2y - 4x) dividing by 2 gives y' = (2y - x)/(y - 2x)
* dividing top an d bottom of fraction by 2
alright. that clears it up. thank you